Applications of Derivatives
Monotonicity and Extrema with Symmetry
nta_pyq_2023_apr
Grade 12
Question:
Let $g(x)=f(x)+f(1-x)$ and $f''(x)>0,\ x\in(0,1)$. If $g$ is decreasing in the interval $(0,\alpha)$ and increasing in the interval $(\alpha,1)$, then $\tan^{-1}(2\alpha)+\tan^{-1}\!\left(\dfrac{1}{\alpha}\right)+\tan^{-1}\!\left(\dfrac{\alpha+1}{\alpha}\right)$ is equal to
\pi
\dfrac{5\pi}{4}
\dfrac{3\pi}{4}
\dfrac{3\pi}{2}
Step-by-Step Solution
Key Concept: $g'(x)=f'(x)-f'(1-x)$. Since $f''(x)>0$, $f'$ is increasing, so $g'(x)=0\Rightarrow x=1-x\Rightarrow x=\tfrac{1}{2}$. Hence $\alpha=\tfrac{1}{2}$.
$\alpha=\frac{1}{2}$. Expression $=\tan^{-1}1+\tan^{-1}2+\tan^{-1}3=\frac{\pi}{4}+\frac{3\pi}{4}=\pi$.
Correct Answer: 1